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Two moles of an ideal monoatomic gas are confined within a cylinder by a massless spring loaded with a frictionless piston of negligible mass and of cross-sectional area 4 xx 10^(-3) m^2. The springis initially in its relaxed state. Now the gas is heated by a heater for some time. During this time the gas expands and does 50 J of work in moving the piston through a distance of 0.1 m. The temperature of the gas increases by 50 K. Calculate the spring constant and the heat supplied by the heater. |
Answer» Solution : A is the INITIAL (equilibrium) position of the piston when the spring is relaxed. When the gas is heated, it expands and pushes the piston up by a distance, say,x. The spring is COMPRESSED, if K is the force constant of the spring and A the area of cross-section of the piston (which is equal to the cross-sectional area of the cylinder),the force in the spring is F = kxand the pressure exerted on the gas by the spring is `P_S=F/A(kx)/A` If `P_0` is the atmospheric pressure, at equilibrium of piston the pressure of the gas in the cylinder is `P=P_0+P_S=P_0+(kx)/A` The INCREASE in the volume of the gas by infinitesimal movement dx of the piston is dV=Adx Thus work done is given as `W=intPdV=int_0^x(P_0+(kx)/A)Adx` `=P_0Aint_0^x dx+k int_0^x xdx` or `W=P_0Ax+1/2kx^2` Given `A=4xx10^(-3)m^2`,x=0.1 m , W=50 J.The atmospheric pressure`P_0=0.76` mof Hg = 0.76 x 9.8 x 13600 = `1.013xx10^5 Nm^(-2)`. Using these values in aboveexpression ofwork and solvingfor k, we get `k=1896 Nm^(-1)` To find heat energyQ suppliedby the heater , we use the firstlaw of thermodynamics. `Q=DeltaU+W` Now `DeltaU=f/2nRDeltaT=nC_VDeltaT` `DeltaU=3/2nRDeltaT` `=3/2xx2xx8.31xx50`=1246.5 J[ For a MONOATOMIC gas we have `C_V=3/2R` ] Thuswe have ,heat suppliedgiven as Q=1246.5+50=1296.5 J |
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